RMO Part-10!

1)Let A=$$\large(a^{2}+4ab+b^{2})$$, where a and b are postive integers.

Prove that A≠2015. (HINT-Use Modular Arithmetic or Try to use the concept of Odd-Even numbers)

2)Prove that any prime number, $$\large(2^{2^n}+1)$$ cannot be represented as a difference of 2 fifth powers of integers.(HINT-Expand the expression and You will get something in common)

3)Find all pairs (x,y) where x and y are integers such that $$\large(x^{3}+11^{3}=y^{3})$$.(HINT-Try to apply identities)

4)If $$\large(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c})$$

Then Prove that,

$$\large(\frac{1}{a^{3}}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^{3}+b^{3}+c^{3}})$$.(HINT-Factorize )

5)Prove that $$\large(2222^{5555}+5555^{2222})$$ is divisible by 7.(HINT-Use Modular Arithmetic or Apply the concept of $$a^{n}±b^{n}$$.)

Today I have posted some easy problems ,So I'm sure that you will be able to solve all these 5 problems in 1.5 hours.

Also try my Set RMO.

Note by Naitik Sanghavi
2 years, 7 months ago

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Here's my solution to (1).

Assume that ,

$$\large (a+b)^{2}+2ab=2015$$

$$\large (a+b)^{2}=2015-2ab$$

So, $$\large (a+b)^{2}$$=Odd

So,$$\large (a+b)$$=Odd(2q+1)For some positive integer q.

So,$$\large 2ab=4p$$For some integer p.

So, $$\large (2q+1)^{2}+4p=2015$$

$$\large 4(q^{2}+q+p)=2014$$

$$\large q^{2}+q+p=\frac{2014}{4}$$

Now we can see that R.H.S.is not an integer but In L.H.S. all are intgers .Thus it does not satisfy the given conditions.Hence Proved.

- 2 years, 6 months ago

4) let $$p(x) = x^3 + lx^2 + mx + p$$ with roots a,b,c then $$p = mn$$ and $$c = -b$$ . Now it easy

- 2 years, 6 months ago

Nice!!You got a different to solve this!

- 2 years, 6 months ago

5) $$5555=4 \pmod 7$$ or $$5555^{2222}=4^{2222} \pmod 7$$

$$4^3=1 \pmod 7$$ or $$4^{2220}=1 \pmod 7$$ or $$4^{2222}=16=2 \pmod 7$$

Therefore, $$5555^{2222}=2 \pmod 7$$.

Similarly,

$$2222=3 \pmod 7$$ or $$2222^{5555}=3^{5555} \pmod 7$$

$$3^3=-1 \pmod 7$$ or $$3^{5553}=(-1)^{1851}=-1=6 \pmod 7$$ or $$3^{5555}=54=5 \pmod 7$$

Therefore, $$5555^{2222}=5 \pmod 7$$

Therefore $$5555^{2222}+2222^{5555}=2+5=7=0 \pmod 7$$

- 2 years, 6 months ago

It seems that you have solved all problems of this note,You can my other problems of this set.Try my set Olympiad Preparation Set for an easier level !!

- 2 years, 6 months ago

2) Suppose $$2^{2^n}+1=a^5-b^5$$,where $$a^5-b^5$$ be the difference of two fifth powers.$$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)\ \quad where \quad a>b$$.

Since $$2^{2^n}+1 \quad is \quad prime, \quad a-b=1$$.

By Fermat's Little theorem, $$2^{2^n}+1=a^5-b^5=a-b \pmod 5=1 \pmod 5$$.

Thus $$2^{2^n}+1$$ cannot be represented as a difference of two fifth powers.

Moreover, $$2^{2^n}+1$$ cannot be represented as the difference of two pth powers, where p is an odd prime.

- 2 years, 6 months ago

1) $$a^2 + 4ab + b^2 = 0, 2 or 1 mod4$$ but 2015 = 3 mod4. A contradiction.

- 2 years, 6 months ago

I seems that you have solved all problems on this Set,You can try my other problems of this set and Try my set Olympiad Preparation setfor an easier level!!

- 2 years, 6 months ago

Just a small correction, $$a^2+4ab+b^2$$ can also be $$0 \pmod 4$$.Try using \pmod for mod. I too did it exactly the same way.

- 2 years, 6 months ago

Yes,of course!! You can post your solutions if you have solved any or have an different solution from others!!!

- 2 years, 6 months ago

Nice Observation!! I'm having an different solution!!You can try out other problems!!I will post my solutions after some days!!!!!!!

- 2 years, 6 months ago

3) Is (-11,0),(0,11) the only solutions?

- 2 years, 6 months ago

Yes!!Correct but you need to prove that there are no other solutions too!!

- 2 years, 6 months ago

@naitik sanghavi $$y= \sqrt [3]{x^3+11^3}=\sqrt [3]{(x+11)(x^2-11x+121)}$$ which is only a perfect cube when either of the terms are zero. Hence, getting the above solution.

- 2 years, 6 months ago

4) This one is quite long! After a lot of solving I got c=-a or c=-b . Substitute both the values and you get the second equation. Am I missing an easier method?

- 2 years, 6 months ago

Here's my solution to (4). $$\large \frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}$$

$$\large \frac{a+b}{ab}=\frac{c-(a+b+c)}{(a+b+c)×c}$$

$$\large (a+b)×c×(a+b+c)=-(a+b)×ab$$ (Don't remove (a+b) from both sides!Why?See Ahead!!!)

$$\large (a+b)×c×(a+b+c)+(a+b)ab=0$$

$$\large (a+b)[c×(a+c)+b×(a+c)]=0$$

$$\large (a+b)(b+c)(c+a)=0$$

a=-b or b=-c or c=-a.

Now substitute anyone value and it will proved!!

- 2 years, 6 months ago

@naitik sanghavi Nice!!.I multiplied (ab+bc+ca) with (a+b+c),did a little bit of factoring and got $$c^2+(a+b)c+ab=0$$ and finally solved the quadratic equation to get the above solution.Your solution is much faster upvoted!!

- 2 years, 6 months ago

You can also generalize it for powers of n and then by induction it will be proved for 3.

- 2 years, 6 months ago

Thanks for your contribution for the Brilliant RMO community :)

- 2 years, 7 months ago

Its OK .I am also preparing for RMO so just posting some problems daily..!!!!

- 2 years, 7 months ago